Question: $f(x, y, z) = (\cos(z), \sin(x + y), x)$ $\text{curl}(f) = $ $\hat{\imath} + $ $\hat{\jmath} + $ $\hat{k}$
$f(x, y, z) = (f_0, f_1, f_2)$ The curl of $f$ : $\begin{aligned} \text{curl}(f) &= \det \begin{bmatrix} {\hat{\imath}} & \hat{\jmath} & \hat{k} \\ \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \\ f_0 & f_1 & f_2 \end{bmatrix} \\ \\ &= \left( \dfrac{\partial f_2}{\partial y} - \dfrac{\partial f_1}{\partial z} \right) \hat{\imath} \\ \\ &+ \left( \dfrac{\partial f_0}{\partial z} - \dfrac{\partial f_2}{\partial x} \right) \hat{\jmath} \\ \\ &+ \left( \dfrac{\partial f_1}{\partial x} - \dfrac{\partial f_0}{\partial y} \right) \hat{k} \end{aligned}$ $\begin{aligned} f_0(x, y, z) &= \cos(z) \\ \\ f_1(x, y, z) &= \sin(x + y) \\ \\ f_2(x, y, z) &= x \end{aligned}$ Let's calculate all the partial derivatives we'll need. $f_0$ $f_1$ $f_2$ $\dfrac{\partial}{\partial x}$ $\cos(x + y)$ $1$ $\dfrac{\partial}{\partial y}$ $0$ $0$ $\dfrac{\partial}{\partial z}$ $-\sin(z)$ $0$ Now we can put it all together. $\begin{aligned} \text{curl}(f) &= \left( \dfrac{\partial f_2}{\partial y} - \dfrac{\partial f_1}{\partial z} \right) \hat{\imath} \\ \\ &+ \left( \dfrac{\partial f_0}{\partial z} - \dfrac{\partial f_2}{\partial x} \right) \hat{\jmath} \\ \\ &+ \left( \dfrac{\partial f_1}{\partial x} - \dfrac{\partial f_0}{\partial y} \right) \hat{k} \\ \\ &= (0 - 0) \hat{\imath} + (- \sin(z) - 1) \hat{\jmath} + (\cos(x + y) - 0) \hat{k} \\ \\ &= 0 \hat{\imath} + (-\sin(z) - 1) \hat{\jmath} + \cos(x + y) \hat{k} \end{aligned}$ In conclusion: $\text{curl}(f) = 0 \hat{\imath} + (-\sin(z) - 1) \hat{\jmath} + \cos(x + y) \hat{k}$